A thin equiconvex lens made of glass of refractive index `3//2` and of focal length `0.3m` in air is sealed into an opening at one end of a tank filled with water `(mu=(3)/(2))`. On the opposite side of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis as shown in Figure. The separation between the lens and mirror is `0.8m. ` A small object is placed outside the tank in front of the lens at a distance oa `0.9m` from the lens along its axis. Find the position (relative to lens) of the image of the object formed by the system.

Answer

`90 cm` from the lens towards right

Explanation

Form lens Maker's formula,
`(1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))`
we have `(1)/(0.3)=((30)/(2)-1)((1)/(R)-(1)/(-R))`
`(Here R_(1)=R_(2)=-R) therefore R=0.4 m`
Now applying `(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)` at air galss surface, we get
`((3)/(2))/v_(1)-(1)/(-(0.9))=((3)/(2)-1)/(0.3) therefore v_(1)=2.7 m`
`i.e,.` first iamge `_(1)` will be formed at `2.7 m` from the lens. This will act as the virtual object for glass water surface. Therefore, applying `(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)` at glass water surface,
We have `(4//3)/(v^(2))-(3//2)/(2.7)=(4//3-3//2)/(-0.3)`
`therefore v_(2)=1.2m`
`i.e,.` second iamge `_(2)` is formed at `1.2 m` from the lens or `0.4 m` from the plane mirror. This will act as virtual object for mirror. Therefore, third real iamge `_(3)` will be formed at a distance of `0.4 m` in front of the mirror after reflection form it. Now this iamge will work as a real object for water-glass interface. Hence, applying
`(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R) "We get" ((3)/(2))/(v_(4))-((4)/(3))/(-(0.8-0.4))=((3)/(2)-(4)/(3))/(0.3)`
`therefore v_(4)=-0.54 m`
`i.e.,` fourth iamge is `0.9 m` relative to the lens (rightward) or the image is formed `0.1 m` behind the mirror .